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3 years ago

If a tree of 5.5 meters casts a shadow of 3 meters. How high will a tower that gives a shadow of 15 meters be?

Physics

1 answer:

n200080 [17]3 years ago

3 0

Answer:

The height of a tower would be 27.5 meters

Explanation:

Notice that in both cases you are dealing with diagrams representing vertical objects and shadows created by the sun, against the ground, so we are in the presence of right angle triangles which have in common another angle (that made by the rays of the sun. So we can use similar triangles and the proportion that is generated based on the quotient of similar sides to find the unknown height (h).

$\frac{5.5}{3} =\frac{h}{15} \\h=\frac{(5.5)\,(15)}{3} \\h=27.5\,\,m$

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If the cold temperature reservoir of a Carnot engine is held at a constant 306 K, what temperature should the hot reservoir be k

Paraphin [41]

Efficiency η of a Carnot engine is defined to be:
η = 1 - Tc / Th = (Th - Tc) / Th 
where 
Tc is the absolute temperature of the cold reservoir, and 
Th is the absolute temperature of the hot reservoir. 

In this case, given is η=22% and Th - Tc = 75K 
Notice that although temperature difference is given in °C it has same numerical value in Kelvins because magnitude of the degree Celsius is exactly equal to that of the Kelvin (the difference between two scales is only in their starting points). 

Th = (Th - Tc) / η 
Th = 75 / 0.22 = 341 K (rounded to closest number) 
Tc = Th - 75 = 266 K 

Lower temperature is Tc = 266 K 
Higher temperature is Th = 341 K

6 0

3 years ago

A 3.0 kg object is loaded into a toy spring gun, and the spring has a spring constant 750N/m. The object compresses the spring b

kiruha [24]

Answer:

Explanation:

mass of object, m = 3 kg

spring constant, K = 750 n/m

compression, x = 8 cm = 0.08 m

angle of gun, θ = 30°

(a) As the ball is launched, it has some velocity due to the compression in the spring, so it has some kinetic energy.

(b) Let v be th evelocity of ball at the tim eof launch.

by using the conservation of energy

1/2 Kx² = 1/2 mv²

750 x 0.08 x 0.08 = 3 x v²

v = 1.265 m/s

By use of the formula of maximum height

$h = \frac{v^{2}Sin^{2}\theta}{2g}$

$h = \frac{1.265^{2}Sin^{2}30}{2\times 9.8}$

h = 0.02 m

h = 2 cm

4 0

3 years ago

Consider a system two point charges. One has charge +q at (x, y,z) -(a,0,0) and another of charge-q at (x, y, z) = (-a, 0,0). 5.

olga2289 [7]

Answer:

electricfield at (0,0,0) is Et = 2 k q / a²

Explanation:

For the first part see the diagram , the field lines start from the positive charge and reach the negative charge, notice that no line should cross, some lines go to infinity

For the second part we use that the electric field is a vector quantity and therefore we add the field of each charge, using the equation

E = k q / r²

Point (0,0,0)

We calculate for the charge -q which is at a distance R = a

E1 = k (-q) / a²

E1 = - kq / a²

As the test charge is positive in the field it goes to the left, attractive force

We calculate for the charge that is also at R = a

E2 = k q / a²

This field goes to the left, repulsive force

We find the total electric field

Et = E1 + E2

Et = kq / a² + kq / a²

Et = 2 k q / a²

Point (0,0, R)

We use the same equations, but with another distance, for the charge -q the distance is R = R+a and for the charge + q the distance is R = R-a

E1 = k q / (R + a)²

E2 = kq / (R-a)²

Et = kq [1 / (R + a)² + 1 / (R-a)²]

Et= kq {[(R-a)² + (R + a)²] / [(R + a)² (R-a)²]}

Et= kq {2 (R² + a²) / [(R + a)² (R-a)²]}

If we use the condition that R> a we can despise in the patents "a"

(R² + a²) = R² (1+ a² / R²) ≈ R²

(R + a)² = R² (1 + a / R)² ≈ R²

(R- a)² = R² (1-a / R)² ≈ R²

Substituting in the total electric field

Et = kq {2 R²) / [R²R²]}

Et =kq 2 / R²

7 0

3 years ago

our friend is constructing a balancing display for an art project. She has one rock on the left (ms=2.25 kgms=2.25 kg) and three

Licemer1 [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The torque produced by the pile of rocks is $\tau = 35.63\ N \cdot m$

The distance of the single for equilibrium to occur is $r_s =1.62 \ m$

Explanation:

From the question we are told that

The mass of the left rock is $m_s = 2.25 \ kg$

The mass of the rock on the right $m_p = 10.1 kg$

The distance from fulcrum to the center of the pile of rocks is $r_p = 0.360 \ m$

Generally the torque produced by the pile of rock is mathematically represented as

$\tau = m_p * g * r_p$

Substituting values

$\tau = 10.1 * 9.8 * 0.360$

$\tau = 35.63\ N \cdot m$

Generally we can mathematically evaluated the distance of the the single rock that would put the system in equilibrium as follows

The torque due to the single rock is

$\tau = m_s * g * r_s$

At equilibrium the both torque are equal

$35.63 = m_s * r_s * g$

Making $r_s$ the subject of the formula

$r_s = \frac{35.63 }{m_s * g}$

Substituting values

$r_s = \frac{35.63 }{2.25 * 9.8}$

$r_s =1.62 \ m$

6 0

3 years ago

A 10cm long wire is pulled along a U-shaped conducting rail in a perpendicular magnetic field. The total resistance of the wire

soldi70 [24.7K]

Answer:

a) $v=4.0m/s$

b) $B=2.958T$

Explanation:

From the question we are told that:

Wire Length $l=10cm=>0.10m$

Resistance $R=0.35$

Force $F=1.0N$

Power $P=4.0W$

Generally the equation for Power is mathematically given by

$P=Fv$

Therefore

$v=\frac{P}{F}$

$v=\frac{4.0}{1.0}$

$v=4.0m/s$

Generally the equation for Magnetic Field is mathematically given by

$B=\frac{\sqrt{PR}}{vl}$

$B=\frac{\sqrt{4*0.35}}{4*0.10}$

$B=2.958T$

5 0

2 years ago